1
500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given: R = 8.3 J K-1 mol-1
- 368 K and 500 J
- 348 K and 300 J
- 378 K and 300 J
- 378 K and 500 J
Answer
(2) 348 K and 300 J
Solution
Argon is monoatomic: Cv = 32R, Cp = 52R = 52 8.3 = 20.75 J K-1 mol-1
Heat is absorbed at constant pressure (1.00 atm): qp = nCp T
T = 5000.5 20.75 48 K
T2 298 + 48 346 K, i.e. about 348 K (the nearest value given)
U = nCv T = qp CvCp = 500 35 = 300 J
(The remaining 200 J is used as expansion work, w = -nR T = -200 J.)
Answer: (2)
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